如何遍历xml
xml文件如下，不知怎么遍历出所有值，求教

<?xml version="1.0" encoding="gb2312"?>
<userinfo>
<clas uid="1" uno="101" uname="a" utype="1">
<guid gname="wx" gtype="3" />
<home addr="" mobil="" mail=""/>
<friendm>
<friendm fnum="" fname="" />
</friendm>
<friendwm fnum="" fname=""/>
</clas>
<clas uid="2" uno="102" uname="b" utype="1">
<guid gname="we" gtype="3" />
<home addr="" mobil="" mail=""/>
<friendm>
<friendm fnum="" fname="" />
</friendm>
<friendwm fnum="" fname=""/>
</clas>
<clas uid="3" uno="1011" uname="c" utype="1">
<guid gname="zf" gtype="3" />
<home addr="" mobil="" mail=""/>
<friendm>
<friendm fnum="" fname="" />
</friendm>
<friendwm fnum="" fname=""/>
</clas>
</userinfo>
[解决办法]
把你问题的标题输入google都有答案。如果你明知这一点，那么你无非就是想找个苦力而已了。
[解决办法]

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
using System.Xml;

namespace ProjectTool
{
    /// <summary>
    /// xml文件操作类
    /// </summary>
    class XmlHelper
    {
        XmlDocument xmlDoc = new XmlDocument();

        #region 获取指定目录下所有子节点的值
        /// <summary>  
        /// 获取指定目录下所有子节点的值  
        /// </summary>  
        /// <param name="strFilePath">文件路径</param>  
        /// <param name="nodeDir">节点目录</param>  
        /// <returns></returns>  
        public Hashtable GetNodeList(string strFilePath, string nodeDir)
        { 
 
            Hashtable strNodeList = new Hashtable();
            try
            {
                XmlDocument xmlDoc = new XmlDocument();
                xmlDoc.Load(strFilePath);


                XmlNodeList nodeList = xmlDoc.SelectSingleNode(nodeDir).ChildNodes;//获取bookstore节点的所有子节点   


                foreach (XmlNode xn in nodeList)     //遍历所有子节点   
                {
                    XmlElement xe = (XmlElement)xn;  //将子节点类型转换为XmlElement类型   
                    strNodeList.Add(xe.GetAttribute("id").ToString(), xe.InnerText.Trim());
                }


            }
            catch (Exception)
            {


                throw;
            }
            return strNodeList;
        }
        #endregion

        #region 获取指定节点的值
        /// <summary>  
        /// 获取指定节点的值  
        /// </summary>  
        /// <param name="strFilePath">文件路径</param>  
        /// <param name="nodeName">节点名称</param>  
        /// <param name="value">设置后的值</param>  
        /// <param name="nodeDir">指定节点所在的节点目录</param>  
        /// <returns></returns>  
        public string GetNodeValue(string strFilePath, string nodeName, string nodeDir)
        {
            string value = null;
            try
            {
                XmlDocument xmlDoc = new XmlDocument();
                xmlDoc.Load(strFilePath); 
 


                XmlNodeList nodeList = xmlDoc.SelectSingleNode(nodeDir).ChildNodes;//获取bookstore节点的所有子节点   


                foreach (XmlNode xn in nodeList)    //遍历所有子节点   
                {
                    XmlElement xe = (XmlElement)xn;  //将子节点类型转换为XmlElement类型   


                    if (xe.Name == nodeName)
                    {
                        value = xe.InnerText.Trim();


                        break;
                    }
                }
            }
            catch (Exception exp)
            {
                throw exp;
            }
            return value;
        }
        #endregion

        #region 获取指定节点下面对应属性的值
        /// <summary>  
        /// 获取指定节点下面对应属性的值  
        /// </summary>  
        /// <param name="strFileName">文件路径</param>  
        /// <param name="nodeName">节点名称</param>  
        /// <param name="nodeDir">指定节点所在的节点目录</param>  
        /// <param name="attribute">节点对应的属性名称</param>  
        /// <returns></returns>  
        public string GetNodeValue(string strFileName, string nodeName, string nodeDir, string attribute)
        {
            string value = null;
            try
            {
                XmlDocument xmlDoc = new XmlDocument();
                xmlDoc.Load(strFileName); 
 
                XmlNodeList nodeList = xmlDoc.SelectSingleNode(nodeDir).ChildNodes;//获取bookstore节点的所有子节点
                foreach (XmlNode xn in nodeList)    //遍历所有子节点   
                {
                    XmlElement xe = (XmlElement)xn;  //将子节点类型转换为XmlElement类型
                    if (xe.Name == nodeName)
                    {
                        //value = xe.InnerText.Trim();  
                        value = (xe).Attributes[attribute].Value;
                        break;
                    }
                }
            }
            catch (Exception exp)
            {
                throw exp;
            }
            return value;
        }
        #endregion
    }
}

XElement xmlDoc = XElement.Parse(xmlString); 

var queryResults = from element in xmlDoc.Elements("clas") 
where element.Element("UID").Value == "3" 
select element; 

xd.Load(@xmlpath);
XmlElement xmlRoot = xd.DocumentElement;
XmlNode node = xmlRoot.SelectSingleNode("userinfo");
foreach (XmlNode xn in node.ChildNodes)
{
if (xn.InnerText == "clas")//循环所有的clas
{
if (xn.Attributes.Count > 0)
{
for (int i = 0; i < xn.Attributes.Count; i++)//循环属性
{
if (xn.Attributes[i].Name == "uid")
Response.Write("属性值:" + xn.Attributes[i].Value + "</br>");
if (xn.Attributes[i].Name == "uno")
Response.Write("属性值:" + xn.Attributes[i].Value + "</br>");
if (xn.Attributes[i].Name == "uname")
Response.Write("属性值:" + xn.Attributes[i].Value + "</br>");
}
}
}
}
[解决办法]

你加好root以后 在把我现在发的代码试试就可以了
另外一种思路
或者你直接从根目录读取也行
[解决办法]

}
}
}
}
[解决办法]
谢谢谢谢谢谢谢谢
[解决办法]

if (xa.Attributes[i].Name == "gtype")
Response.Write("属性值:" + xa.Attributes[i].Value + "</br>");
}
}
}
}
if (xn.Attributes.Count > 0)
{
for (int i = 0; i < xn.Attributes.Count; i++)//循环属性
{
if (xn.Attributes[i].Name == "uid")
Response.Write("属性值:" + xn.Attributes[i].Value + "</br>");
if (xn.Attributes[i].Name == "uno")
Response.Write("属性值:" + xn.Attributes[i].Value + "</br>");
if (xn.Attributes[i].Name == "uname")
Response.Write("属性值:" + xn.Attributes[i].Value + "</br>");
}
}
}
}
我这样处理也行，
另外一种思路： 如果你循环的次数多的话 用递归去处理更方便
------解决方案--------------------

}
}
}
if (xn.Attributes.Count > 0)
{
for (int i = 0; i < xn.Attributes.Count; i++)//循环属性
{
if (xn.Attributes[i].Name == "uid")
Response.Write("属性值:" + xn.Attributes[i].Value + "</br>");
if (xn.Attributes[i].Name == "uno")
Response.Write("属性值:" + xn.Attributes[i].Value + "</br>");
if (xn.Attributes[i].Name == "uname")
Response.Write("属性值:" + xn.Attributes[i].Value + "</br>");
}
}
}
}
我这样处理也行，
另外一种思路： 如果你循环的次数多的话 用递归去处理更方便
[解决办法]
我觉得你可以去看下XML序列化和反序列化方面的资料，如果你的XML文件中的数据很多的话，用序列化会比你遍历节点快一些
[解决办法]

            string xml = @"<userinfo> 
  <clas uid=""1"">
 <uinfo uno=""101"" uname=""a"" utype=""1"" addr="""" mobil="""" mail="""" fruum=""3""/>
      </clas>  
  <clas uid=""2"">
 <uinfo uno=""102"" uname=""b"" utype=""1"" addr="""" mobil="""" mail="""" fruum=""4""/>
      </clas>  
 <clas uid=""3"">
 <uinfo uno=""103"" uname=""c"" utype=""1"" addr="""" mobil="""" mail="""" fruum=""5""/>
      </clas>  

 </userinfo> ";
            DataSet ds = new DataSet();
            using (System.IO.StringReader reader = new System.IO.StringReader(xml))
            {
                try
                {
                    System.Xml.XmlTextReader xmlReader = new System.Xml.XmlTextReader(reader);
                    ds.ReadXml(xmlReader);
                    xmlReader.Close();
                }
                finally
                {
                    reader.Close();
                }
            }